Let P(n) be 3 + 3 * 5 + 3 * 52+...+3 * 5n = 3(5n + 1 -1)/4 whenever n is a nonne

Question

Let P(n) be 3 + 3 * 5 + 3 * 52+...+3 * 5n = 3(5n + 1 -1)/4 whenever n is a nonnegative P(0) is true since 3 * 5 = 3 = 3(5 0 + 1 -1)/4 Assume that P(n) is true. Then 3 + 3 * 5 + 3 * 52 +...+3 * 5n + 3 * 5 n + 1 = 3(5 n + 1 -1)/4 + 3 * 5n + 1 (3(5n + 1 -1) + 3 * 4 * 5n + 1)/4 = 3(5n + 2 -1) 3(5(n + 1) + 1 -1)/4 Hence P(n + 1) is true. Let P(n) be "1 + 3 +...+ (2n + 1) = (n + 1)(2n + 1)(2n + 3)/3 whenever n is a nonnegative P(0) is true since 1 = 1 = (0 + 1)(2 * 0 + 1)(2 * 0 + 1)(2 * 0 + 3)/3. Assume that P(n) is true. Then 1 + 3 +... + (2n + 1) + (2(n + 1) + 1)

Explanation / Answer

Basically what they are doing is proving it for P(n+1) given that P(n) is true

Now P(n+1) = 3 + 3*5 + 3*5^2 + ....+ 3* 5^n + 3* 5^(n+1)

Now you have assumed P(n) is true

then 3 + 3*5 + 3*5^2 + ... + 3 * 5^n = 3* (5^(n+1) - 1) /4

replace this in the previous equation

so you get 3* (5^(n+1) - 1) /4 + 3* 5^(n+1)

now take 3 common, then you get

3 { (5^(n+1) -1) / 4 + 5^(n+1) }

then make the denominator 4 common

= 3 { (5^(n+1) - 1 + 4 * 5^(n+1) )/4 }

take 5 ^ (n+1) common

then

3 { 5 * 5 ^ (n+1) -1} /4

= 3 { 5^ (n+2) - 1} /4

which is P(n+1) right hand side .

Hence by the method of induction it is proved.

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